Răspuns :
non coliniare daca det (ABC ) ≠ 0
5 2 1
-3 7 1 = 35 + 15 + 2 - 7 + 6 + 25 ≠ 0
1 -5 1
⇒ A , B si C nu sunt coliniare
5 2 1
-3 7 1 = 35 + 15 + 2 - 7 + 6 + 25 ≠ 0
1 -5 1
⇒ A , B si C nu sunt coliniare
[tex] \left[\begin{array}{ccc}1&1&1\\x1&x2&x3\\y1&y2&y3\end{array}\right] [/tex]
[tex] \left[\begin{array}{ccc}1&1&1\\5&-3&1\\2&7&-5\end{array}\right] = 15+2+35-(-6+7-25)=15+2+35+6-7+25=76 \neq 0[/tex]
[tex] \left[\begin{array}{ccc}1&1&1\\5&-3&1\\2&7&-5\end{array}\right] = 15+2+35-(-6+7-25)=15+2+35+6-7+25=76 \neq 0[/tex]